For a simple counterexample, consider $X =[0,1]$. If the map $t \mapsto \delta_t$ were differentiable in either sense then for every bounded linear functional $F$ on $AE(X)$ the map $t\mapsto F(\delta_t)$ would be differentiable. Recalling that the dual of $AE(X)$ is ${\rm Lip}_0(X)$, differentiability at $t$ would imply that every Lipschitz function on $[0,1]$ is differentiable at $t$, which is obviously false.

This fails for $X = \mathbb{R}$, and hence for every nonzero Banach space, since they all contain copies of $\mathbb{R}$. If the map $t \mapsto \delta_t$ were differentiable in either sense then for every bounded linear functional $F$ on $AE(\mathbb{R})$ the map $t\mapsto F(\delta_t)$ would be differentiable. Recalling that the dual of $AE(\mathbb{R})$ is ${\rm Lip}_0(\mathbb{R})$, differentiability at $t$ would imply that every Lipschitz function on $\mathbb{R}$ is differentiable at $t$, which is obviously false.